Characterization of variational symmetries I have the following definition of variational symmetry, this is from Bruce Van Brunt's Calculus of Variations book
Suppose we have a functional of the form
$$
J(y)= \int_{x_{0}}^{x_{1}} f(x,y,y')dx
$$
and a smooth transformation $X=\theta (x,y: \epsilon)$ , $Y=\psi (x,y: \epsilon)$ where $\epsilon$ is a parameter.
Definition. The functional $J$ is invariant under the above transformation if, for all $\epsilon$ sufficiently small, in any subinterval $[a,b] \subset [x_{0},x_{1}]$ we have that
$$
\int_{a}^{b} f(x,y,y')dx= \displaystyle \int_{a_{\epsilon}}^{b_{\epsilon}} f(X,Y,Y')dX
$$
for all smooth functions $y$ defined on $[a,b]$. Here, $a_{\epsilon}=\theta (a,y(a): \epsilon)$, $b_{\epsilon}=\theta (b,y(b): \epsilon)$
And, $ \epsilon$ is a parameter such that $\theta (x,y;  0)=x$ and $\psi (x,y;  0)=y$.
Once said that, I would appreciate your help for a rigorous proof of the converse of the following theorem

Theorem 9.4.1. Let
$$
J(y)= \int_{x_{0}}^{x_{1}} f(x,y,y')dx.
$$
The transformation above with infinitesimal generators $\xi$ and $\eta$ is a variational symmetry for $J$ if and only if
$$
\xi \displaystyle \frac{\partial f}{\partial x} + \eta \frac{\partial f}{\partial y} + ( \eta ' - y' \xi ' ) \frac{\partial f}{\partial y'} + \xi ' f=0 \tag{9.38}
$$
for any smooth function $y$ on $ [x_{0},x_{1}] $.
Here,
$$
 \xi (x,y)=  \frac{\partial \theta}{ \partial \epsilon }   \Big |_{(x,y;0)}^{},
$$
$$
 \eta (x,y)=  \frac{\partial \psi}{ \partial \epsilon }   \Big |_{(x,y;0)}^{}.
$$

I have been thinking about it , but, I believe that, at least under this definition of invariance the converse is false. The author refers us to the book Introduction to the calculus of variations and its applications by Frederic Wan for a proof, however, I see nowhere that the converse is proved.
Any help?
Remark:
The reason for which I think the converse it's false is because if we consider the functional
$$
\displaystyle \int_{x_{0}}^{x_{1}} xy'^2 dx
$$
And the transformation
$$
X=x+2 \epsilon x ln(x), Y=(1+ \epsilon ) y
$$
Then, equation $\xi  \frac{\partial f}{\partial x} + \eta \frac{\partial f}{\partial y} + ( \eta ' - y' \xi ' ) \frac{\partial f}{\partial y'} + \xi ' f=0$ holds , but, I think there is no way that the above transformation is a variational symmetry since
$$
XY ' ^2 dX= \left( xy'^2+ \frac{(x \epsilon ^2+ 2 \epsilon ^3 x ln(x))y'^2}{(1+2 \epsilon ln(x)+ 2 \epsilon)}   \right) dx
$$
And this , definitely does not agree with the definition. Or am I wrong? If so, why? :(
Update: Here are the computations that shows equation $\xi  \frac{\partial f}{\partial x} + \eta \frac{\partial f}{\partial y} + ( \eta ' - y' \xi ' ) \frac{\partial f}{\partial y'} + \xi ' f=0$ holds:
We have that $\xi=2xln(x)$ and $\eta=y$, so
$$\xi f_{x}+ \eta f_{y}+(\eta' - y' \xi ') f_{y'}+ \xi' f = 2x ln(x) y'^2 + (y'-y'(2ln(x)+2))2xy'+(2ln(x)+2)xy'^2=2xln(x)y'^2-2xy'^2-4y'^2xln(x)+2xy'^2ln(x)+2xy'^2=0$$
 A: The converse is essentially what you can get from Picard's theorem for ODE.
The counter example is not working. In your case $\xi=2x\ln x$ and $\eta=y$.
So you have $2x\ln x\cdot y'^2+ y\cdot 0 + (0-y'(2\ln x + 2))\cdot 2xy'+ 0$, which doesn't seem to be 0.
Update: I was wrong in interpreting the conditions, the OP is right, the condition will take $\eta$'s dependence on x through y in the evaluation of $\eta'$, which should have been defined as $\eta'(x,y)=\frac{\partial\eta}{\partial x}+\frac{\partial\eta}{\partial y} y'$.
So the counterexample stands, and the condition would only work if $\eta_\epsilon(x,y)$ and $\xi_\epsilon(x,y)$ are defined for a small neighborhood of $\epsilon=0$ with the above condition. That's also why the above ends up being almost true up to a $O(\epsilon^2)$ term, since the condition is satisfied up to a deviation for $\eta_\epsilon$ and $\xi_\epsilon$ away from $\epsilon=0$.
Update again: My guess is that the condition was actually to use $(\xi,\eta)$ as a 2D vector field, as the generator of diffeomorphism on the 2D plane, and  compute $(\theta,\psi)=\exp((\xi,\eta)\epsilon).$
I figured the exponential map for OP's example can be worked out in closed form. So here it is:
$X=x^{e^{2\epsilon}},\quad Y=y e^\epsilon.$
Now $\frac{\partial X}{\partial x} = e^{2\epsilon}\frac{X}{x},$ and $Y'=\frac{\partial Y}{\partial y} \frac{d y}{d x} \frac{\partial x}{\partial X}=e^{\epsilon} y' \frac{x}{X} e^{-2\epsilon},$ which lead to $XY^{'2}dX=xy^{'2}dx$ precisely.
A: *

*It seems that OP is pondering the following question:

Does the off-shell Noether identity$^1$ (9.38) imply that the functional $J(\epsilon)$ has a 1-parameter off-shell symmetry for sufficiently small finite $\epsilon>0$?

This is of course not necessarily the case. We can always change the transformations for $X$ and $Y$ at order ${\cal O}(\epsilon^2)$ in a manner that breaks the finite symmetry of $J(\epsilon)$ without altering eq. (9.38).


*In contrast the correct statement is:

The off-shell Noether identity (9.38) holds iff there is an infinitesimal off-shell symmetry
$$\left.\frac{dJ(\epsilon)}{d\epsilon}\right|_{\epsilon=0}~=~0.$$

References:

*

*B. van Brunt,  The Calculus of Variations, 2004; p. 213.


*F. Wan, Intro to the Calculus of Variations and its Applications, 1995; p. 189.
--
$^1$ The following is not part of OP's question, but if the reader  wonders what the off-shell Noether identity (9.38) has to do with Noether's theorem, let us mention that it can be rewritten as
$$\begin{align} 0~=~& \eta \frac{\partial f}{\partial y} + ( \dot{\eta} -  \dot{\xi} \dot{y}) \frac{\partial f}{\partial \dot{y}} + \dot{\xi}f +\xi  \frac{\partial f}{\partial x} \cr
~=~&\eta_0 \frac{\partial f}{\partial y} +  \dot{\eta}_0  \frac{\partial f}{\partial \dot{y}} + \frac{d(\xi f)}{dx} \cr
~=~&\eta_0 \underbrace{\frac{\delta J}{\delta y}}_{\text{EL eq.}} + \dot{Q} ,\end{align}\tag{9.38} $$
where $$Q~=~\eta_0\frac{\partial f}{\partial \dot{y}}+\xi f$$
is the (bare) Noether charge; where
$$\eta_0~=~\eta-\xi\dot{y}$$
is the vertical generator; and where dot means a total derivative $\frac{d}{dx}$. In particular, we deduce that the Noether charge $Q$ is conserved on-shell.
